Internal problem ID [1359]
Internal file name [OUTPUT/1360_Sunday_June_05_2022_02_13_02_AM_25472708/index.tex]
Book: Elementary differential equations with boundary value problems. William F. Trench.
Brooks/Cole 2001
Section: Chapter 7 Series Solutions of Linear Second Equations. 7.6 THE METHOD OF
FROBENIUS II. Exercises 7.6. Page 374
Problem number: 7.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second order series method. Regular singular point. Repeated root"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {x^{2} y^{\prime \prime }+x \left (x^{2}+x +1\right ) y^{\prime }+x \left (2-x \right ) y=0} \] With the expansion point for the power series method at \(x = 0\).
The ODE is \[ x^{2} y^{\prime \prime }+\left (x^{3}+x^{2}+x \right ) y^{\prime }+\left (-x^{2}+2 x \right ) y = 0 \] Or \[ x \left (y^{\prime } x^{2}+x y^{\prime \prime }+x y^{\prime }-y x +y^{\prime }+2 y\right ) = 0 \] For \(x \neq 0\) the above simplifies to \[ x y^{\prime \prime }+\left (x^{2}+x +1\right ) y^{\prime }-y \left (x -2\right ) = 0 \] The type of the expansion point is first determined. This is done on the homogeneous part of the ODE. \[ x^{2} y^{\prime \prime }+\left (x^{3}+x^{2}+x \right ) y^{\prime }+\left (-x^{2}+2 x \right ) y = 0 \] The following is summary of singularities for the above ode. Writing the ode as \begin {align*} y^{\prime \prime }+p(x) y^{\prime } + q(x) y &=0 \end {align*}
Where \begin {align*} p(x) &= \frac {x^{2}+x +1}{x}\\ q(x) &= -\frac {x -2}{x}\\ \end {align*}
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([0, \infty , -\infty ]\)
Irregular singular points : \([\infty ]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ x^{2} y^{\prime \prime }+\left (x^{3}+x^{2}+x \right ) y^{\prime }+\left (-x^{2}+2 x \right ) y = 0 \] Let the solution be represented as Frobenius power series of the form \[ y = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r} \] Then \begin{align*} y^{\prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1} \\ y^{\prime \prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2} \\ \end{align*} Substituting the above back into the ode gives \begin{equation} \tag{1} x^{2} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right )+\left (x^{3}+x^{2}+x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (-x^{2}+2 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) = 0 \end{equation} Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r +2} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n +r +2} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 x^{1+n +r} a_{n}\right ) = 0 \end{equation} The next step is to make all powers of \(x\) be \(n +r\) in each summation term. Going over each summation term above with power of \(x\) in it which is not already \(x^{n +r}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r +2} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}a_{n -2} \left (n +r -2\right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} \left (n +r -1\right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n +r +2} a_{n}\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (-a_{n -2} x^{n +r}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}2 x^{1+n +r} a_{n} &= \moverset {\infty }{\munderset {n =1}{\sum }}2 a_{n -1} x^{n +r} \\ \end{align*} Substituting all the above in Eq (2A) gives the following equation where now all powers of \(x\) are the same and equal to \(n +r\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}a_{n -2} \left (n +r -2\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} \left (n +r -1\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =2}{\sum }}\left (-a_{n -2} x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}2 a_{n -1} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+x^{n +r} a_{n} \left (n +r \right ) = 0 \] When \(n = 0\) the above becomes \[ x^{r} a_{0} r \left (-1+r \right )+x^{r} a_{0} r = 0 \] Or \[ \left (x^{r} r \left (-1+r \right )+x^{r} r \right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ x^{r} r^{2} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ r^{2} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 0\\ r_2 &= 0 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ x^{r} r^{2} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([0, 0]\).
Since the root of the indicial equation is repeated, then we can construct two linearly independent solutions. The first solution has the form \begin {align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\tag {1A} \end {align*}
Now the second solution \(y_{2}\) is found using \begin {align*} y_{2}\left (x \right ) &= y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n} x^{n +r}\right )\tag {1B} \end {align*}
Then the general solution will be \[ y = c_{1} y_{1}\left (x \right )+c_{2} y_{2}\left (x \right ) \] In Eq (1B) the sum starts from 1 and not zero. In Eq (1A), \(a_{0}\) is never zero, and is arbitrary and is typically taken as \(a_{0} = 1\), and \(\{c_{1}, c_{2}\}\) are two arbitray constants of integration which can be found from initial conditions. We start by finding the first solution \(y_{1}\left (x \right )\). Eq (2B) derived above is now used to find all \(a_{n}\) coefficients. The case \(n = 0\) is skipped since it was used to find the roots of the indicial equation. \(a_{0}\) is arbitrary and taken as \(a_{0} = 1\). Substituting \(n = 1\) in Eq. (2B) gives \[ a_{1} = \frac {-2-r}{\left (1+r \right )^{2}} \] For \(2\le n\) the recursive equation is \begin{equation} \tag{3} a_{n} \left (n +r \right ) \left (n +r -1\right )+a_{n -2} \left (n +r -2\right )+a_{n -1} \left (n +r -1\right )+a_{n} \left (n +r \right )-a_{n -2}+2 a_{n -1} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {n a_{n -2}+n a_{n -1}+r a_{n -2}+r a_{n -1}-3 a_{n -2}+a_{n -1}}{n^{2}+2 n r +r^{2}}\tag {4} \] Which for the root \(r = 0\) becomes \[ a_{n} = \frac {\left (-a_{n -2}-a_{n -1}\right ) n +3 a_{n -2}-a_{n -1}}{n^{2}}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = 0\) and after as more terms are found using the above recursive equation.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-2-r}{\left (1+r \right )^{2}}\) | \(-2\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {-r^{3}+6 r +7}{\left (1+r \right )^{2} \left (r +2\right )^{2}} \] Which for the root \(r = 0\) becomes \[ a_{2}={\frac {7}{4}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-2-r}{\left (1+r \right )^{2}}\) | \(-2\) |
| \(a_{2}\) | \(\frac {-r^{3}+6 r +7}{\left (1+r \right )^{2} \left (r +2\right )^{2}}\) | \(\frac {7}{4}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {2 r^{4}+10 r^{3}+6 r^{2}-23 r -28}{\left (1+r \right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2}} \] Which for the root \(r = 0\) becomes \[ a_{3}=-{\frac {7}{9}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-2-r}{\left (1+r \right )^{2}}\) | \(-2\) |
| \(a_{2}\) | \(\frac {-r^{3}+6 r +7}{\left (1+r \right )^{2} \left (r +2\right )^{2}}\) | \(\frac {7}{4}\) |
| \(a_{3}\) | \(\frac {2 r^{4}+10 r^{3}+6 r^{2}-23 r -28}{\left (1+r \right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2}}\) | \(-{\frac {7}{9}}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {r^{6}+5 r^{5}-11 r^{4}-96 r^{3}-146 r^{2}-16 r +77}{\left (1+r \right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (4+r \right )^{2}} \] Which for the root \(r = 0\) becomes \[ a_{4}={\frac {77}{576}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-2-r}{\left (1+r \right )^{2}}\) | \(-2\) |
| \(a_{2}\) | \(\frac {-r^{3}+6 r +7}{\left (1+r \right )^{2} \left (r +2\right )^{2}}\) | \(\frac {7}{4}\) |
| \(a_{3}\) | \(\frac {2 r^{4}+10 r^{3}+6 r^{2}-23 r -28}{\left (1+r \right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2}}\) | \(-{\frac {7}{9}}\) |
| \(a_{4}\) | \(\frac {r^{6}+5 r^{5}-11 r^{4}-96 r^{3}-146 r^{2}-16 r +77}{\left (1+r \right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (4+r \right )^{2}}\) | \(\frac {77}{576}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {-3 r^{7}-41 r^{6}-189 r^{5}-259 r^{4}+468 r^{3}+1716 r^{2}+1651 r +434}{\left (1+r \right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (4+r \right )^{2} \left (5+r \right )^{2}} \] Which for the root \(r = 0\) becomes \[ a_{5}={\frac {217}{7200}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-2-r}{\left (1+r \right )^{2}}\) | \(-2\) |
| \(a_{2}\) | \(\frac {-r^{3}+6 r +7}{\left (1+r \right )^{2} \left (r +2\right )^{2}}\) | \(\frac {7}{4}\) |
| \(a_{3}\) | \(\frac {2 r^{4}+10 r^{3}+6 r^{2}-23 r -28}{\left (1+r \right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2}}\) | \(-{\frac {7}{9}}\) |
| \(a_{4}\) | \(\frac {r^{6}+5 r^{5}-11 r^{4}-96 r^{3}-146 r^{2}-16 r +77}{\left (1+r \right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (4+r \right )^{2}}\) | \(\frac {77}{576}\) |
| \(a_{5}\) | \(\frac {-3 r^{7}-41 r^{6}-189 r^{5}-259 r^{4}+468 r^{3}+1716 r^{2}+1651 r +434}{\left (1+r \right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (4+r \right )^{2} \left (5+r \right )^{2}}\) | \(\frac {217}{7200}\) |
For \(n = 6\), using the above recursive equation gives \[ a_{6}=\frac {-r^{9}-15 r^{8}-47 r^{7}+365 r^{6}+3206 r^{5}+9364 r^{4}+10369 r^{3}-2834 r^{2}-15026 r -8813}{\left (1+r \right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (4+r \right )^{2} \left (5+r \right )^{2} \left (6+r \right )^{2}} \] Which for the root \(r = 0\) becomes \[ a_{6}=-{\frac {8813}{518400}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-2-r}{\left (1+r \right )^{2}}\) | \(-2\) |
| \(a_{2}\) | \(\frac {-r^{3}+6 r +7}{\left (1+r \right )^{2} \left (r +2\right )^{2}}\) | \(\frac {7}{4}\) |
| \(a_{3}\) | \(\frac {2 r^{4}+10 r^{3}+6 r^{2}-23 r -28}{\left (1+r \right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2}}\) | \(-{\frac {7}{9}}\) |
| \(a_{4}\) | \(\frac {r^{6}+5 r^{5}-11 r^{4}-96 r^{3}-146 r^{2}-16 r +77}{\left (1+r \right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (4+r \right )^{2}}\) | \(\frac {77}{576}\) |
| \(a_{5}\) | \(\frac {-3 r^{7}-41 r^{6}-189 r^{5}-259 r^{4}+468 r^{3}+1716 r^{2}+1651 r +434}{\left (1+r \right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (4+r \right )^{2} \left (5+r \right )^{2}}\) | \(\frac {217}{7200}\) |
| \(a_{6}\) | \(\frac {-r^{9}-15 r^{8}-47 r^{7}+365 r^{6}+3206 r^{5}+9364 r^{4}+10369 r^{3}-2834 r^{2}-15026 r -8813}{\left (1+r \right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (4+r \right )^{2} \left (5+r \right )^{2} \left (6+r \right )^{2}}\) | \(-{\frac {8813}{518400}}\) |
For \(n = 7\), using the above recursive equation gives \[ a_{7}=\frac {4 r^{10}+112 r^{9}+1264 r^{8}+7170 r^{7}+19330 r^{6}+4756 r^{5}-116404 r^{4}-318504 r^{3}-355034 r^{2}-145179 r +8008}{\left (1+r \right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (4+r \right )^{2} \left (5+r \right )^{2} \left (6+r \right )^{2} \left (r +7\right )^{2}} \] Which for the root \(r = 0\) becomes \[ a_{7}={\frac {143}{453600}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-2-r}{\left (1+r \right )^{2}}\) | \(-2\) |
| \(a_{2}\) | \(\frac {-r^{3}+6 r +7}{\left (1+r \right )^{2} \left (r +2\right )^{2}}\) | \(\frac {7}{4}\) |
| \(a_{3}\) | \(\frac {2 r^{4}+10 r^{3}+6 r^{2}-23 r -28}{\left (1+r \right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2}}\) | \(-{\frac {7}{9}}\) |
| \(a_{4}\) | \(\frac {r^{6}+5 r^{5}-11 r^{4}-96 r^{3}-146 r^{2}-16 r +77}{\left (1+r \right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (4+r \right )^{2}}\) | \(\frac {77}{576}\) |
| \(a_{5}\) | \(\frac {-3 r^{7}-41 r^{6}-189 r^{5}-259 r^{4}+468 r^{3}+1716 r^{2}+1651 r +434}{\left (1+r \right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (4+r \right )^{2} \left (5+r \right )^{2}}\) | \(\frac {217}{7200}\) |
| \(a_{6}\) | \(\frac {-r^{9}-15 r^{8}-47 r^{7}+365 r^{6}+3206 r^{5}+9364 r^{4}+10369 r^{3}-2834 r^{2}-15026 r -8813}{\left (1+r \right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (4+r \right )^{2} \left (5+r \right )^{2} \left (6+r \right )^{2}}\) | \(-{\frac {8813}{518400}}\) |
| \(a_{7}\) | \(\frac {4 r^{10}+112 r^{9}+1264 r^{8}+7170 r^{7}+19330 r^{6}+4756 r^{5}-116404 r^{4}-318504 r^{3}-355034 r^{2}-145179 r +8008}{\left (1+r \right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (4+r \right )^{2} \left (5+r \right )^{2} \left (6+r \right )^{2} \left (r +7\right )^{2}}\) | \(\frac {143}{453600}\) |
Using the above table, then the first solution \(y_{1}\left (x \right )\) becomes \begin{align*} y_{1}\left (x \right )&= a_{0}+a_{1} x +a_{2} x^{2}+a_{3} x^{3}+a_{4} x^{4}+a_{5} x^{5}+a_{6} x^{6}+a_{7} x^{7}+a_{8} x^{8}\dots \\ &= 1-2 x +\frac {7 x^{2}}{4}-\frac {7 x^{3}}{9}+\frac {77 x^{4}}{576}+\frac {217 x^{5}}{7200}-\frac {8813 x^{6}}{518400}+\frac {143 x^{7}}{453600}+O\left (x^{8}\right ) \\ \end{align*} Now the second solution is found. The second solution is given by \[ y_{2}\left (x \right ) = y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n} x^{n +r}\right ) \] Where \(b_{n}\) is found using \[ b_{n} = \frac {d}{d r}a_{n ,r} \] And the above is then evaluated at \(r = 0\). The above table for \(a_{n ,r}\) is used for this purpose. Computing the derivatives gives the following table
| \(n\) | \(b_{n ,r}\) | \(a_{n}\) | \(b_{n ,r} = \frac {d}{d r}a_{n ,r}\) | \(b_{n}\left (r =0\right )\) |
| \(b_{0}\) | \(1\) | \(1\) | N/A since \(b_{n}\) starts from 1 | N/A |
| \(b_{1}\) | \(\frac {-2-r}{\left (1+r \right )^{2}}\) | \(-2\) | \(\frac {r +3}{\left (1+r \right )^{3}}\) | \(3\) |
| \(b_{2}\) | \(\frac {-r^{3}+6 r +7}{\left (1+r \right )^{2} \left (r +2\right )^{2}}\) | \(\frac {7}{4}\) | \(\frac {r^{4}-3 r^{3}-24 r^{2}-46 r -30}{\left (r +2\right )^{3} \left (1+r \right )^{3}}\) | \(-{\frac {15}{4}}\) |
| \(b_{3}\) | \(\frac {2 r^{4}+10 r^{3}+6 r^{2}-23 r -28}{\left (1+r \right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2}}\) | \(-{\frac {7}{9}}\) | \(\frac {-4 r^{6}-30 r^{5}-40 r^{4}+201 r^{3}+762 r^{2}+997 r +478}{\left (r +3\right )^{3} \left (r +2\right )^{3} \left (1+r \right )^{3}}\) | \(\frac {239}{108}\) |
| \(b_{4}\) | \(\frac {r^{6}+5 r^{5}-11 r^{4}-96 r^{3}-146 r^{2}-16 r +77}{\left (1+r \right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (4+r \right )^{2}}\) | \(\frac {77}{576}\) | \(\frac {-2 r^{9}-15 r^{8}+64 r^{7}+1075 r^{6}+4650 r^{5}+8812 r^{4}+4548 r^{3}-9852 r^{2}-16988 r -8084}{\left (1+r \right )^{3} \left (r +2\right )^{3} \left (r +3\right )^{3} \left (4+r \right )^{3}}\) | \(-{\frac {2021}{3456}}\) |
| \(b_{5}\) | \(\frac {-3 r^{7}-41 r^{6}-189 r^{5}-259 r^{4}+468 r^{3}+1716 r^{2}+1651 r +434}{\left (1+r \right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (4+r \right )^{2} \left (5+r \right )^{2}}\) | \(\frac {217}{7200}\) | \(\frac {9 r^{11}+209 r^{10}+1920 r^{9}+8034 r^{8}+5769 r^{7}-94779 r^{6}-473517 r^{5}-1121767 r^{4}-1522043 r^{3}-1167285 r^{2}-431134 r -39712}{\left (1+r \right )^{3} \left (r +2\right )^{3} \left (r +3\right )^{3} \left (4+r \right )^{3} \left (5+r \right )^{3}}\) | \(-{\frac {1241}{54000}}\) |
| \(b_{6}\) | \(\frac {-r^{9}-15 r^{8}-47 r^{7}+365 r^{6}+3206 r^{5}+9364 r^{4}+10369 r^{3}-2834 r^{2}-15026 r -8813}{\left (1+r \right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (4+r \right )^{2} \left (5+r \right )^{2} \left (6+r \right )^{2}}\) | \(-{\frac {8813}{518400}}\) | \(\frac {3 r^{14}+81 r^{13}+690 r^{12}-1434 r^{11}-75047 r^{10}-683625 r^{9}-3350559 r^{8}-9722533 r^{7}-14651453 r^{6}+1600677 r^{5}+56328076 r^{4}+122022818 r^{3}+134469042 r^{2}+79674152 r +20273544}{\left (1+r \right )^{3} \left (r +2\right )^{3} \left (r +3\right )^{3} \left (4+r \right )^{3} \left (5+r \right )^{3} \left (6+r \right )^{3}}\) | \(\frac {93859}{1728000}\) |
| \(b_{7}\) | \(\frac {4 r^{10}+112 r^{9}+1264 r^{8}+7170 r^{7}+19330 r^{6}+4756 r^{5}-116404 r^{4}-318504 r^{3}-355034 r^{2}-145179 r +8008}{\left (1+r \right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (4+r \right )^{2} \left (5+r \right )^{2} \left (6+r \right )^{2} \left (r +7\right )^{2}}\) | \(\frac {143}{453600}\) | \(\frac {-16 r^{16}-784 r^{15}-16992 r^{14}-212142 r^{13}-1644632 r^{12}-7626912 r^{11}-13834248 r^{10}+71523314 r^{9}+668108016 r^{8}+2726754259 r^{7}+6982218308 r^{6}+11832092046 r^{5}+12839081488 r^{4}+7603715879 r^{3}+578454492 r^{2}-2102187772 r -940999248}{\left (1+r \right )^{3} \left (r +2\right )^{3} \left (r +3\right )^{3} \left (4+r \right )^{3} \left (5+r \right )^{3} \left (6+r \right )^{3} \left (r +7\right )^{3}}\) | \(-{\frac {311177}{42336000}}\) |
The above table gives all values of \(b_{n}\) needed. Hence the second solution is \begin{align*} y_{2}\left (x \right )&=y_{1}\left (x \right ) \ln \left (x \right )+b_{0}+b_{1} x +b_{2} x^{2}+b_{3} x^{3}+b_{4} x^{4}+b_{5} x^{5}+b_{6} x^{6}+b_{7} x^{7}+b_{8} x^{8}\dots \\ &= \left (1-2 x +\frac {7 x^{2}}{4}-\frac {7 x^{3}}{9}+\frac {77 x^{4}}{576}+\frac {217 x^{5}}{7200}-\frac {8813 x^{6}}{518400}+\frac {143 x^{7}}{453600}+O\left (x^{8}\right )\right ) \ln \left (x \right )+3 x -\frac {15 x^{2}}{4}+\frac {239 x^{3}}{108}-\frac {2021 x^{4}}{3456}-\frac {1241 x^{5}}{54000}+\frac {93859 x^{6}}{1728000}-\frac {311177 x^{7}}{42336000}+O\left (x^{8}\right ) \\ \end{align*} Therefore the homogeneous solution is \begin{align*} y_h(x) &= c_{1} y_{1}\left (x \right )+c_{2} y_{2}\left (x \right ) \\ &= c_{1} \left (1-2 x +\frac {7 x^{2}}{4}-\frac {7 x^{3}}{9}+\frac {77 x^{4}}{576}+\frac {217 x^{5}}{7200}-\frac {8813 x^{6}}{518400}+\frac {143 x^{7}}{453600}+O\left (x^{8}\right )\right ) + c_{2} \left (\left (1-2 x +\frac {7 x^{2}}{4}-\frac {7 x^{3}}{9}+\frac {77 x^{4}}{576}+\frac {217 x^{5}}{7200}-\frac {8813 x^{6}}{518400}+\frac {143 x^{7}}{453600}+O\left (x^{8}\right )\right ) \ln \left (x \right )+3 x -\frac {15 x^{2}}{4}+\frac {239 x^{3}}{108}-\frac {2021 x^{4}}{3456}-\frac {1241 x^{5}}{54000}+\frac {93859 x^{6}}{1728000}-\frac {311177 x^{7}}{42336000}+O\left (x^{8}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} \left (1-2 x +\frac {7 x^{2}}{4}-\frac {7 x^{3}}{9}+\frac {77 x^{4}}{576}+\frac {217 x^{5}}{7200}-\frac {8813 x^{6}}{518400}+\frac {143 x^{7}}{453600}+O\left (x^{8}\right )\right )+c_{2} \left (\left (1-2 x +\frac {7 x^{2}}{4}-\frac {7 x^{3}}{9}+\frac {77 x^{4}}{576}+\frac {217 x^{5}}{7200}-\frac {8813 x^{6}}{518400}+\frac {143 x^{7}}{453600}+O\left (x^{8}\right )\right ) \ln \left (x \right )+3 x -\frac {15 x^{2}}{4}+\frac {239 x^{3}}{108}-\frac {2021 x^{4}}{3456}-\frac {1241 x^{5}}{54000}+\frac {93859 x^{6}}{1728000}-\frac {311177 x^{7}}{42336000}+O\left (x^{8}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \left (1-2 x +\frac {7 x^{2}}{4}-\frac {7 x^{3}}{9}+\frac {77 x^{4}}{576}+\frac {217 x^{5}}{7200}-\frac {8813 x^{6}}{518400}+\frac {143 x^{7}}{453600}+O\left (x^{8}\right )\right )+c_{2} \left (\left (1-2 x +\frac {7 x^{2}}{4}-\frac {7 x^{3}}{9}+\frac {77 x^{4}}{576}+\frac {217 x^{5}}{7200}-\frac {8813 x^{6}}{518400}+\frac {143 x^{7}}{453600}+O\left (x^{8}\right )\right ) \ln \left (x \right )+3 x -\frac {15 x^{2}}{4}+\frac {239 x^{3}}{108}-\frac {2021 x^{4}}{3456}-\frac {1241 x^{5}}{54000}+\frac {93859 x^{6}}{1728000}-\frac {311177 x^{7}}{42336000}+O\left (x^{8}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} \left (1-2 x +\frac {7 x^{2}}{4}-\frac {7 x^{3}}{9}+\frac {77 x^{4}}{576}+\frac {217 x^{5}}{7200}-\frac {8813 x^{6}}{518400}+\frac {143 x^{7}}{453600}+O\left (x^{8}\right )\right )+c_{2} \left (\left (1-2 x +\frac {7 x^{2}}{4}-\frac {7 x^{3}}{9}+\frac {77 x^{4}}{576}+\frac {217 x^{5}}{7200}-\frac {8813 x^{6}}{518400}+\frac {143 x^{7}}{453600}+O\left (x^{8}\right )\right ) \ln \left (x \right )+3 x -\frac {15 x^{2}}{4}+\frac {239 x^{3}}{108}-\frac {2021 x^{4}}{3456}-\frac {1241 x^{5}}{54000}+\frac {93859 x^{6}}{1728000}-\frac {311177 x^{7}}{42336000}+O\left (x^{8}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \left (\frac {d}{d x}y^{\prime }\right )+\left (x^{3}+x^{2}+x \right ) y^{\prime }+\left (-x^{2}+2 x \right ) y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}y^{\prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }=\frac {\left (x -2\right ) y}{x}-\frac {\left (x^{2}+x +1\right ) y^{\prime }}{x} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }+\frac {\left (x^{2}+x +1\right ) y^{\prime }}{x}-\frac {\left (x -2\right ) y}{x}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=\frac {x^{2}+x +1}{x}, P_{3}\left (x \right )=-\frac {x -2}{x}\right ] \\ {} & \circ & x \cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x \cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=1 \\ {} & \circ & x^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=0 \\ {} & \circ & x =0\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & x \left (\frac {d}{d x}y^{\prime }\right )+\left (x^{2}+x +1\right ) y^{\prime }+\left (2-x \right ) y=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \\ {} & {} & y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot y\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & x^{m}\cdot y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r +m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -m \\ {} & {} & x^{m}\cdot y=\moverset {\infty }{\munderset {k =m}{\sum }}a_{k -m} x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot y^{\prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..2 \\ {} & {} & x^{m}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) x^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & x^{m}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +r \right ) x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x \cdot \left (\frac {d}{d x}y^{\prime }\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x \cdot \left (\frac {d}{d x}y^{\prime }\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) x^{k +r -1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & x \cdot \left (\frac {d}{d x}y^{\prime }\right )=\moverset {\infty }{\munderset {k =-1}{\sum }}a_{k +1} \left (k +1+r \right ) \left (k +r \right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} r^{2} x^{-1+r}+\left (a_{1} \left (1+r \right )^{2}+a_{0} \left (2+r \right )\right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k +1} \left (k +1+r \right )^{2}+a_{k} \left (k +r +2\right )+a_{k -1} \left (k -2+r \right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r^{2}=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r =0 \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & a_{1} \left (1+r \right )^{2}+a_{0} \left (2+r \right )=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & k^{2} a_{k +1}+\left (a_{k}+a_{k -1}+2 a_{k +1}\right ) k +2 a_{k}-2 a_{k -1}+a_{k +1}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & \left (k +1\right )^{2} a_{k +2}+\left (a_{k +1}+a_{k}+2 a_{k +2}\right ) \left (k +1\right )+2 a_{k +1}-2 a_{k}+a_{k +2}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {k a_{k}+k a_{k +1}-a_{k}+3 a_{k +1}}{k^{2}+4 k +4} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=-\frac {k a_{k}+k a_{k +1}-a_{k}+3 a_{k +1}}{k^{2}+4 k +4} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}, a_{k +2}=-\frac {k a_{k}+k a_{k +1}-a_{k}+3 a_{k +1}}{k^{2}+4 k +4}, a_{1}+2 a_{0}=0\right ] \end {array} \]
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying a quadrature checking if the LODE has constant coefficients checking if the LODE is of Euler type trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Trying a Liouvillian solution using Kovacics algorithm <- No Liouvillian solutions exists -> Trying a solution in terms of special functions: -> Bessel -> elliptic -> Legendre -> Kummer -> hyper3: Equivalence to 1F1 under a power @ Moebius -> hypergeometric -> heuristic approach -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius -> Mathieu -> Equivalence to the rational form of Mathieu ODE under a power @ Moebius trying a solution in terms of MeijerG functions -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius <- Heun successful: received ODE is equivalent to the HeunB ODE, case c = 0 `
✓ Solution by Maple
Time used: 0.031 (sec). Leaf size: 71
Order:=8; dsolve(x^2*diff(y(x),x$2)+x*(1+x+x^2)*diff(y(x),x)+x*(2-x)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \left (c_{2} \ln \left (x \right )+c_{1} \right ) \left (1-2 x +\frac {7}{4} x^{2}-\frac {7}{9} x^{3}+\frac {77}{576} x^{4}+\frac {217}{7200} x^{5}-\frac {8813}{518400} x^{6}+\frac {143}{453600} x^{7}+\operatorname {O}\left (x^{8}\right )\right )+\left (3 x -\frac {15}{4} x^{2}+\frac {239}{108} x^{3}-\frac {2021}{3456} x^{4}-\frac {1241}{54000} x^{5}+\frac {93859}{1728000} x^{6}-\frac {311177}{42336000} x^{7}+\operatorname {O}\left (x^{8}\right )\right ) c_{2} \]
✓ Solution by Mathematica
Time used: 0.008 (sec). Leaf size: 153
AsymptoticDSolveValue[x^2*y''[x]+x*(1+x+x^2)*y'[x]+x*(2-x)*y[x]==0,y[x],{x,0,7}]
\[ y(x)\to c_1 \left (\frac {143 x^7}{453600}-\frac {8813 x^6}{518400}+\frac {217 x^5}{7200}+\frac {77 x^4}{576}-\frac {7 x^3}{9}+\frac {7 x^2}{4}-2 x+1\right )+c_2 \left (-\frac {311177 x^7}{42336000}+\frac {93859 x^6}{1728000}-\frac {1241 x^5}{54000}-\frac {2021 x^4}{3456}+\frac {239 x^3}{108}-\frac {15 x^2}{4}+\left (\frac {143 x^7}{453600}-\frac {8813 x^6}{518400}+\frac {217 x^5}{7200}+\frac {77 x^4}{576}-\frac {7 x^3}{9}+\frac {7 x^2}{4}-2 x+1\right ) \log (x)+3 x\right ) \]